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First one deleted.

Please put it on a new post.

 

Find the constant b such that \(\left(5x^2-3x+\frac{7}{3}\right)(ax^2+bx+c) = 15x^4 - 14x^3 + 20x^2 - \frac{25}{3}x + \frac{14}{3}\)

 May 17, 2019
edited by Melody  May 18, 2019
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Please repost the first one on  a seperate thread.     One question per post.

 

2nd one:

 

Find the constant b such that 

 

\(\left(5x^2-3x+\frac{7}{3}\right)(ax^2+bx+c) = 15x^4 - 14x^3 + 20x^2 - \frac{25}{3}x + \frac{14}{3}\\ 5a=15 \qquad a=3\qquad so\\ \left(5x^2-3x+\frac{7}{3}\right)(3x^2+bx+c) = 15x^4 - 14x^3 + 20x^2 - \frac{25}{3}x + \frac{14}{3}\\ \frac{7c}{3}=\frac{14}{3}\qquad c=2 \qquad so\\ \left(5x^2-3x+\frac{7}{3}\right)(3x^2+bx+2) = 15x^4 - 14x^3 + 20x^2 - \frac{25}{3}x + \frac{14}{3}\\~\\ (5b-9)x^3=-14x^3\\ 5b-9=-14\\ 5b=-5\\~\\ {b=-1} \\~\\ \left(5x^2-3x+\frac{7}{3}\right)(3x^2-1x+2) = 15x^4 - 14x^3 + 20x^2 - \frac{25}{3}x + \frac{14}{3}\\~\\ check\\ (-5-9)=-14 \quad and \quad(10+3+7)=20 \quad and \quad (-6-\frac{7}{3})=-\frac{25}{3}\\ great\)

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 May 18, 2019

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